In an underdetermined linear regression where the parameters to estimate outnumber the observations, why is it the case the residual sum of squares, defined as:

$$ RSS = \sum_{i=1}^n (y_i-\widehat{y}_i)^2 $$

equal zero?

  • $\begingroup$ one way to look at it: if you have at least as many $\beta_p$ as $y_i$, through a change of basis you can make it so that each $x_i$ has zeros in all entries except the $i$th one such that each $y_i$ can have its own regression coefficient. $\endgroup$ Commented Feb 6, 2023 at 3:41
  • $\begingroup$ @JohnMadden What about the situation I describe in my answer? // If you believe the situation I describe is contained in your approach, please do post as an answer. While I do not believe your approach to be 100% correct (because of something annoying like I describe in my answer), I had never thought about it that for for the case where an over-parameterized regression can get perfect (in-sample) predictions. $\endgroup$
    – Dave
    Commented Feb 6, 2023 at 3:48
  • 2
    $\begingroup$ @Dave I didn't mention the case with duplicate X's because you did :) you're right, in that case there's no way to rewrite the x's as I describe: rather, we can think of those two observations as sharing a regression coefficient. $\endgroup$ Commented Feb 6, 2023 at 3:51
  • $\begingroup$ In geometry one used to talk about points in general position $\endgroup$ Commented Feb 6, 2023 at 4:58
  • $\begingroup$ "In an underdetermined linear regression setup, why does the residual sum of squares equal 0?" The question is based in a false premise. $\endgroup$ Commented Feb 6, 2023 at 16:36

3 Answers 3


That need not be the case.

It is routine to say that a polynomial of degree greater than the sample size will give a perfect fit to the data. However, this will not be the case if two points have equal feature ($X$) values but unequal outcome ($y$) values. No function can accommodate that. At least one of those points will have a fitted value $\hat y_i$ not equal to its observed $y_i$ value, thus a strictly positive squared residual.


For example...

X  Y
1  1
2  2
3  3
3  4
4  5
5  6

No matter what a model predicts for x = 3, that model will make a prediction that is incorrect, resulting in a positive sum of squared residuals.

This really has nothing to do with $p>n$. If you have two instances with equal feature values and unequal outcomes, the model cannot ever get both right.

x1 x2 x3 x4 x5 x6  y
0  0  0  0  0  0  1.1
0  0  1  1  1  0  0.8
0  0  0  0  1  1  0.3
0  1  0  1  0  0  1.1
0  1  0  1  0  0  0.5

The final two rows have identical feature values yet different outcomes, so a model will get at least one of those wrong, leading to a positive sum of squared residuals.

  • $\begingroup$ Your new example doesn't satisfy $n < p$? $\endgroup$
    – Zhanxiong
    Commented Feb 8, 2023 at 17:48

With one predictor (and an intercept): two points determines a lines exactly, with less than two points there are infinitely many lines that pass through that point ...

With two predictors (and an intercept): three points determines a plane exactly, with less than three points infinitely many planes will fit perfectly ...

The same principle continues into higher dimension.

  • 1
    $\begingroup$ (some exceptions as noted in the answer by @Dave) $\endgroup$ Commented Feb 6, 2023 at 3:39

$\DeclareMathOperator{\rank}{rank}$ Suppose all observations of explanatory variables are identical, then it can be shown that \begin{align} RSS = \sum_{i = 1}^n(y_i - \bar{y})^2. \tag{1} \end{align} Hence $RSS > 0$ unless $y_1, \ldots, y_n$ are all identical.

The reason of this example cannot provide a perfect fit is that the rank of the design matrix $X \in \mathbb{R}^{n \times p}$ is $1$. Following this key observation, there are actually countless examples of $RSS > 0$ as long as $\rank(X) < n$ (unless $y$ is a multiple of $e = 1_n \in \mathbb{R}^n$). This is clear from the projection point of view of least squares, because $\|P_{\text{col}(X)}y\| < \|y\|$ under the condition $\rank(X) < n$ and the linear algebra fact $\dim(\text{col}(X)) = \rank(X) < n$ (there is a theorem in linear algebra states that the rank of a matrix is equal to the dimension of column space).

The conjecture will be true if $\operatorname{rank}(X) = n$, i.e., $X$ is of full row rank. In this case, the dimension of column space is $n$ too, which implies that the columns of $X$ spans the whole space $\mathbb{R}^n$, hence $y$ can be perfectly expressed as a linear combination of columns of $X$ (i.e., $y = P_{\text{col}(X)}y$), this gives $RSS = \|y - P_{\text{col}(X)}y\|^2 = 0$.

Proof of $(1)$. Suppose there are $n$ observations $(y_i, x_{i1}, \ldots, x_{ip}) \equiv (y_i, x_1, \ldots, x_p)$, $i = 1, \ldots, n$. Then the least squares minimizes the objective function \begin{align} \sum_{i = 1}^n (y_i - (\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p))^2. \tag{2} \end{align} Set $t = \beta_0 + \beta_1 x_1 + \cdots + \beta_px_p$, then it is easy to see the $t$ minimizes $(2)$ is $t^* = \bar{y}$. This implies that the minimizer $(\hat{\beta}_0, \ldots, \hat{\beta}_p)$ must satisfy $\hat{\beta}_0 + \hat{\beta}_1 x_1 + \cdots + \hat{\beta}_px_p = \bar{y}$, which is equivalent to $\hat{y}_i = \bar{y}$. This leads to $RSS = \sum_{i = 1}^n (y_i - \bar{y})^2$.

  • $\begingroup$ I don’t agree about me having defined the sum to involve subtracting $\bar y$. Could you please explain your rationale? $\endgroup$
    – Dave
    Commented Feb 8, 2023 at 17:28
  • $\begingroup$ @Dave It can be proved, if the rows of $X$ are identical, the RSS is like this. I will add a proof later (basically all $\hat{y}_i = \bar{y}$). $\endgroup$
    – Zhanxiong
    Commented Feb 8, 2023 at 17:30
  • $\begingroup$ What is the relationship to what I wrote? $\endgroup$
    – Dave
    Commented Feb 8, 2023 at 17:34
  • $\begingroup$ You didn't write. I kind of dug a little deeper to your example. $\endgroup$
    – Zhanxiong
    Commented Feb 8, 2023 at 17:37
  • 1
    $\begingroup$ Sorry, I may have misread your example, I will remove the citation to your answer then. $\endgroup$
    – Zhanxiong
    Commented Feb 8, 2023 at 17:45

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